(C) 4 5 (A) 2 3 (B) 3 4 (D) 1. Sol. Given : The impulse response g(t) of system G. The overall impulse response of two cascaded blocks. i.e.

Transcription

1 One Mark Questions ( 5) Q. The impulse response g(t) of a system G, is as shown in figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in figure (b)? (a) (b) (A) 3 (B) 3 4 (C) 4 5 (D) (D) Sol. Given : The impulse response g(t) of system G. The overall impulse response of two cascaded blocks. h( t) g( t) g( t ) i.e. [ u( t) u( t )] [ u( t) u( t )] r( t) r( t ) r( t ) Therefore, the maximum value attained by the h (t) is. Hence, the correct option is (D) Q. If the sum of the diagonal elements of a matrix is 6, then the maximum possible value of determinant of the matrix is.

2 (9) Q.3 An inductor is connected in parallel with a capacitor as shown in the figure. As the frequency of current i is increased, the impedance (Z ) of the network varies as (A) (B) (C) (D) (B) Sol. Given : Impedance of the circuit, Z X L X C Case-I : If w LC jl jc jl jc jl The impedance Z LC jl LC i.e. Capacitive jl

3 3 Case-II If LC The impedance Case III If w LC Z j L i.e. inductive The impedance Z i.e. open circuit Hence, the correct option is (B) Q.4 Base load power plants are P. Wind farms Q. Run-of-river plants R. Nuclear power plants S. Diesel power plants (A) P, Q and S only (C) P, Q and R only (D) Sol. (B) P, R and S only (D) Q and R only Base load power plants supply power consistently on 4 hour basis. Wind farms, Run-of-rivers plants and Nuclear power plants are base load plants. Hence, the correct option is (C) Q.5 If a continuous function f (x) does not have a root in the interval [a, b], then which one of the following statements is TRUE? (A) f ( a) f ( b) (B) f ( a) f ( b) (C) f ( a) f ( b) (D) (C) f( a) f() b Q.6 A separately excited DC generator has an armature resistance of.ω and negligible armature inductance. At rated field current and rated rotor speed, its open-circuit voltage is V. When this generator is operated at half the rated speed, with half the rated field current, an un-charged μf capacitor is suddenly connected across the armature terminals. Assume that the speed remains unchanged during the transient. At what time (in microsecond) after the capacitor is connected with the voltage across it reach 5 V? (A) 6.5 (B) 69.3 (C) 73.5 (D) 77.3 (B) Sol. Given : A separately excited DC generator Ra., E V DC Case I : At rated speed and rated field current open circuit voltage, E V

4 4 Case II : At Half of rated speed and half field current we know generated voltage ( E) Speed (N) Flux ( ) E E N N N Now, N, E 5 V Time constant RC. 6 4 sec / Voltage built up across capacitor V ( t) E ( e t ) At time t T, V ( t) 5 V 5 5( e ) C T 4 5 5( e ) T 69.3 sec Hence, the correct option is (B) T C Q.7 A (-5 A) moving coil ammeter has a voltage drop of. V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (-5 A) is. (.) Sol.. to.3 Given : A ( 5 A) moving coil ammeter Rm Ammeter resistance Im Full scale deflection current

5 5 Voltage drop I R. V m m Rsh External shunt resistance needed to extend its range to ( 5 A) From figure I R I R sh sh m m ( I I ) R I R m sh m m I R m m Rsh I Im R. m sh Q.8 Consider a function f, r where r is the distance from the origin and r is the unit vector in the r radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is (A) (B) π (C) 4π (D) Rπ (A) Q.9 A random variable X has probability density function f ( x ) as given below : (.5) If the expected value a bx for x f( x) otherwise EX [ ], than P[ X.5] is. 3 Q. Consider a one-turn rectangular loop of wire placed in a uniform magnetic field as shown in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the loop is.4 Ω, and its inductance is negligible. The magnetic flux density (in Tesla) is a function of time, and is given by B(t) =.5 sinωt, where ω = π 5 radian/second. The power absorbed (in Watt) by the loop from the magnetic field is.

6 6 (.6) Q. In the 4 multiplexer, the output F is given by F = A B. Find the required input I3 I I I (A) (B) (C) (D) (B) Sol. Given : F A B F AB AB Also the Boolean expression for 4MUX output F SS I SSI SSI SSI 3 (i) F ABI ABI ABI ABI 3 (ii) Comparing equation (i) and (ii) we get I, I, I, I 3 Hence, the correct option is (B) Q. In the following chopper, the duty ratio of switch S is.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is.

7 7 () Sol. Given : The given chopper is step down chopper Duty ratio.4 Vin V We know, V V Output voltage, V.4 V 8V in Charging current 8 5 I 3 I A Q.3 For the signal-flow graph shown in the figure, which one of the following expressions is equal to the transfer function Ys ()? X s () X ( s ) G (A) G ( G ) G (B) G ( G ) G (C) GG G (D) GG Q.4 In the given circuit, the silicon transistor has β = 75 and a collector voltage Vc 9V and R C is. Then the ratio of R B

8 8 ( to ) Sol. Given : 75, V C 9 V Applying KVL in output loop, we have 5 I ( ) R 9 B C I ( ) R V (i) B Applying KVL in input loop, we have From equation (i) C 5 I ( ) R I R.7 B C B B 5 6 I R.7 Dividing equation (ii) by (i) B B I R 8.3 V (ii) B B IBRB 8.3 I ( ) R 6 B C RB 8.3 (75 ) 5.3 R 6 C Q.5 The self-inductance of the primary winding of a single phase, 5 Hz, transformer is 8 mh, and that of the secondary winding is 6 mh. The mutual inductance between these two windings is 48 mh. The

9 9 secondary winding of this transformer is short circuited and the primary winding is connected to a 5 Hz, single phase, sinusoidal voltage source. The current flowing in both the windings is less than their respective rated currents. The resistance of both windings can be neglected. In this condition, what is the effective inductance (in mh) seen by the source? (A) 46 (B) 44 (C) (D) 9 (A) Sol. Given : Single Phase 5 Hz transformer M 48 mh Leakage inductance of primary, Where, a turn ratio : l Ll L am L mh And secondary leakage inductance Ll a L am 6 48 mh Ll M Effective inductance Ll L M Hence, the correct option is (A) l mh 48 Q.6 Of the four characteristics given below, which are the major requirements for an instrumentation amplifier? P. High common mode rejection ratio Q. High input impedance R. High linearity

10 Sol. S. High output impedance (A) P, Q and R only (C) P, Q and S only (A) Major requirements of an instrumentation amplifier. High common mode rejection ratio. High input impedance 3. High linearity 4. Low output impedance Hence, the correct option is (A) (B) P and R only (D) Q, R and S only Q.7 When the Wheatstone bridge shown in the figure is used to find the value of resistor R X, the galvanometer G indicates zero current when R 5, R 65and R3. If R3 is known with ± 5% tolerance on its nominal value of, what is the range of R X in Ohms? (A) [3.5, 36.5] (B) [5.89, 34.] (C) [7., 43.] (D) [.5, 39.75] (A) Sol. Given : R 5 R 65 R3 i.e. 95 5% tolerance to 5

11 At Balance condition, current I g and R Rx RR3 R RR 3 x R Case I : R3 95 Case II : R R x 5 R 3.5 x 655 R x 5 R 36.5 x Hence, the correct option is (A) Q.8 For the given circuit, the Thevenin equivalent is to be determined. The Thevenin voltage, V Th (in Volt), seen from terminal AB is. (3.3 to 3.4) Sol. Given : For above figure VA VA i i Applying KCL at node A. VA VA VA ( i) From equation (i) i i i ( i) (i)

12 i.6 A Now current through resistance I AB V A ( i) i I.68 A AB The thevenin voltage V TH I AB VTH i V.6 seen from terminal AB Q.9 Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 75 MW from System to System, the voltage at the two ends, and the current are given by: V 5 kv, V 485 kv and I.5 ka. If the direction of power flow is to be reversed (that is, from System to System ) without changing the electrical connections, then which one of the following combinations is feasible? (a) V 5 kv, V 485kV and I.5kA (b) V 485 kv, V 5kV and I.5kA (c) V 5 kv, V 485kV and I.5kA (d) V 5 kv, V 485kV and I.5kA (B) Sol. If the direction of power flow is to be reversed (i.e. from system to system ) without changing electrical connections, then voltage polarities are to be reversed while preserving the direction of current. Hence, the correct option is (B) Q. A Bode magnitude plot for the transfer function G(s) of a plant is shown in the figure. Which one of the following transfer functions best describes the plant?

13 3 ( s ) ( s ) (A) (B) (C) ( s ) s (D) s ss ( ) ss ( ) ( s ) (D) Q. The primary mmf is least affected by the secondary terminal conditions in a (A) Power transformer (B) Potential transformer (C) Current transformer (D) Distribution transformer (B) Sol. In current transformer, primary has single turn and current does not depend on secondary current. Thus primary mmf is not dependent on secondary terminal conditions. Q. The voltages developed across the 3 Ω and Ω resistors shown in the figure are 6 V and V respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5 V voltage source? (A) 5 (B) 7 (C) (D) 4 (A) Sol. Given : Applying KCL, we get

14 4 i x i A x Power delivered by 5 V voltage source P i 5 x 5 P 5W Hence, the correct option is (A) Q.3 Consider the circuit shown in the figure. In this circuit R k, and C F. The input voltage is sinusoidal with a frequency of 5 Hz, represented as a phasor with magnitudev i and phase angle radian as shown in the figure. The output voltage is represented as a phasor with magnitude V and phase angle δ radian. What is the value of the output phase and δ (in radian) relative to the phase angle of the input voltage? (A) (B) (C) (D) Sol. Given : (D) R k, C F Applying KCL at Non-inverting node VNI V VNI R Cs VNI ( VNI V ) Cs R

15 5 VNI Cs VCs R VRCs VNI RCs Applying KCL at Inverting Node VI V VI V R Cs VI V ( VI V ) Cs R V VI Cs VCs R R VRCs V VI RCs RCs According to virtual ground concept V V NI I Equating equation (i) and (ii) V RCs V RCs V RCs RCs RCs V ( V V ) RCs V ( ) ( ) j Vi j RC { Vi Vi V V} (i) (ii) V ViRC (iii) and V V (iv) {Given} Comparing equation (iii) and (iv) we get Hence, the correct option is (D) Q.4 A steady current I is flowing in the x direction through each of two infinitely long wires at shown in the figure. The permeability of the medium is. The B field at (, L, ) is L y as

16 6 4I (A) z ˆ 3L (A) 4I 3 (B) z I ˆ (C) (D) z ˆ 3 L 4L Q.5 A moving average function is given by y ( t t ) ( ). T u d If the input u is a sinusoidal signal of t T frequency Hz, then in steady state, the output y will lag u (in degree) by. T (9) Sol. Given : A moving average function y( t) u( ) d t T t T Let input u ( ) sin, T The output, y( t) u( ) d T t T t T t tt f T sin d t [ cos( )] T t T [cos ( t T ) cos t ] [cos t cos T sin t sin T cos t ] y( t) cos t { T } y t t ( ) sin( 9 )

17 7 The output yt () will lag u by 9 in the steady state Q.6 Two single-phase transformer T and T each rated at 5 kva are operated in parallel. Percentage impedances of T and T are ( + j6) and (.8 + j4.8) respectively. To share a load of kva at.8 lagging power factor, the contribution of T (in kva) is. (555.55) Sol. 554 to 556 Given : Two single phase transformer T and T Percentage impedance z j6 and z.8 j4.8 S load KVA at.8 lagging p.f For parallel operation of T and T The load shared by transformer T is Z S S Z Z l j6 ( ) j6.8 j S KVA at.8 p.f lag Q.7 A parallel plate capacitor is partially filled with glass of dielectric constant 4. as shown below. The dielectric strengths of air and glass are 3 kv/cm and 3 kv/cm, respectively. The maximum voltage (in kilovolts), which can be applied across the capacitor without any breakdown, is (8.75 or 37.5) Q.8 f (A, B, C, D) = ΠM(,, 3, 4, 5, 7, 9,,, 3, 4, 5) is a maxterm representation of Boolean function f (A, B, C, D) where A is the MSB and D is the LSB. The equivalent minimized representation of this function is (A) ( A C D)( A B D) (B) ACD ABD (C) ACD ABCD ABCD (D) ( B C D)( A B C D)( A B C D) (A) Sol. Given :

18 8 f ( A, B, C, D) M (,,3, 4,5,7,9,,,3,4,5) Using K-map Equivalent minimized representation f ( A C D)( A B D) Hence, the correct option is (A) Q.9 A 3-phase 5 Hz square wave (6-step) VSI feeds a 3-phase, 4 pole induction motor. The VSI line voltage has a dominant 5th harmonic component. If the operating slip of the motor with respect to fundamental component voltage is.4, the slip of the motor with respect to 5 th harmonic component of voltage is. (.88) Sol..6 to. Given : 3-phase 4 pole induction motor P 4 Slip of the motor with respect to fundamental component voltage, S.4 Ns Synchronous speed Nr Rotor speed N N ( s) r s 5(.4) N 44 rpm r f 5 5 rpm P 4 The fifth harmonic filed will be rotating in opposite direction to the rotor at 5 times fundamental frequency synchronous speed. Ns rpm Now slip with respect to 5 th harmonic component of voltage S5 75 S5. Q.3 Consider a discrete time signal given by n n x[ n] (.5) u[ n] (.5) u[ n ] The region of convergence of its Z-transform would be

19 9 (A) The region inside the circle of radius.5 and centered at origin (B) The region outside the circle of radius.5 and centered at origin (C) The annular region between the two circles, both centered at origin and having radii.5 and.5. (D) The entire Z plane (C) Sol. Given : n n x[ n] (.5) u[ n] (.5) u[ n ] (i) and x[ n] x[ n] x[ n] (ii) Comparing equation (i) and (ii) we get [ ] (.5) n x [ ] n u n and x [ n] (.5) n u[ n ] Taking z-transform of above functions and n n ( ) [ ] (.5 ) n n X z x n z z a.5 z, ROC : z.5 n n ( ) [ ] (.5 ) n n X z x n z z z, ROC : z.5 z.5 The region of convergence of z-transform of x [n] is.5 z.5 Hence, the correct option is (C) Q.3 The signum function is given by x ; x sgn ( x) x ; x The Fourier series expansion of sgn (cos(t)) has (A) Only sine terms with all harmonics (B) Only cosine terms with all harmonics

20 Sol. (C) Only sine terms with even numbered harmonics (D) Only cosine terms with odd numbered harmonics (D) Given : The signum function is given by The function x, x sgn ( x) x, x cost, cost sgn (cos t) cos t, cost The above function can also be described as sgn (cos t) for cost for cost The functions cos t and sgn (cos t) are as follows The function sgn ( cos t ) is an even function, i.e. f ( t) f ( t) with half wave symmetry i.e. T f () t f t. Thus, the function sgn ( cos t ) has only consine terms due to even symmetry with odd numbered harmonics due to half wave symmetry. Hence, the correct option is (D) Q.3 Find the transfer function Y() s X() s of the system given below

21 Sol. G G (A) HG HG G G (C) H ( G G ) (C) G G (B) HG HG G G (D) H ( G G ) Q.33 A V, 5 Hz, two-winding transformer is rated at kva. Its windings are connected as an autotransformer of rating /6 V. A resistive load of Ω is connected to the high voltage (6 V) side 4 of the auto-transformer. The value of equivalent load resistance (in Ohm) as seen from low voltage side is. (.33) Sol..3 to.4 Given : A /4 V, 5 Hz, two winding transformer rated KVA I AB ICD Full load current Full load current A 5 A 3 4 When two winding transformer is connected as auto transformer. 3

22 I 5 A CD Output KVA of auto transformer 6 5 Input current 3 3 KVA Per unit resistance on output side (HV) Per unit resistance on input side (LV) 6 5 R 5 Since per unit resistance on both side is same R.33 5 R5 6 Q.34 An unbalanced DC Wheatstone bridge is shown in the figure. At what value of p will the magnitude of V be maximum? (A) ( x) (B) ( x) (C) (A) Sol. Given : ( x) (D) ( x)

23 3 From figure, V Va Vb (i) Now using voltage division rule R( x) E( x) Va E R ( x ) PR x P and V R E b E PR R P Putting the values of Va and Vb in equation (i) We have ( x) E V E x P P To get maximum value of P for which V is maximum dv dp Differentiating equation (ii) with respect to P and equating to zero dv dp E( x)() E () ( x P) ( P) ( x) ( x P) ( P) ( x)( P) ( x P) ( x)( P P) ( x) P P( x) ( x) ( x) P P( x) ( x) P P( x) P x P x x ( ) ( ) ( ) P ( x) ( x)( x ) P ( x) x( x) P ( x) P x Hence, the correct option is (A) (ii)

24 4 Q.35 The single-phase full-bridge voltage source inverter (VSI), shown in figure, has an output frequency of 5 Hz. It uses unipolar pulse width modulation with switching frequency of 5 khz and modulation index of.7. For V in = V DC, L = 9.55 mh, C = µf, and R 5, the amplitude of the fundamental component in the output voltage V (in Volt) under steady-state is. 6 to 64 Sol. Given : M a Modulation index =.7 V V dc The rms voltage at any harmonic n is given by 4V dc ( V ) n Ma n For n fundamental Harmonic 4V dc V Ma 4.7 V 63 Q.36 A DC motor has the following specifications: hp, 37.5 A, 3 V; flux/pole =. Wb, number of poles = 4, number of conductors = 666, number of parallel paths =. Armature resistance =.67 Ω. The armature reaction is negligible and rotational losses are 6 W. The motor operates from a 3 V DC supply. If the motor runs at rpm, the output torque produced (in Nm) is. (57 to 58) Sol. Given : A DC motor, hp, 37.5 A, 3 V. I Armature current = 37.5 A a Flux/pole =. Wb P Number of poles = 4 Z Number of conductors = 666 A Number of parallel paths = Ra Armature resistance.67 Rotational loss 6 W P L N Speed = rpm

25 5 Back emf developed by motor PZ N Eb A Volt V Eb Now, Ia R Ia a A Power developed EI b a W Power output = Power developed Rotational losses W N Angular velocity rad/ sec. Power output Output torque T Q.37 Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is (A) 5 (B) (C) 7 (D) 6 3 (D) Q.38 Sol. 5.7 to 6. Given :

26 6 The inertia constant Mechanical input power Initial rotor angle H MJ / MVA 5 P pu From dynamics of synchronous machine moment of inertia M Consider swing equation d Pm Pe M dt Pe m d where, angular acceleration dt due to three phase fault M M 45 s 45elect deg / The change in rotor angle after. sec t M GH where, G pu 8 f 45 (.).9 new Q.39 A 5 Hz generating unit has H-constant of MJ/MVA. The machine is initially operating in steady state at synchronous speed, and producing pu of real power. The initial value of the rotor angle δ is 5,when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at pu, the value of δ in degrees,. second after the fault is. (5.9) Q.39 Determine the correctness or otherwise of the following Assertion [A] and the Reason [R]. Assertion : Fast decoupled load flow method gives approximate load flow solution because it uses several assumptions. Reason : Accuracy depends on the power mismatch vector tolerance. (A) Both [A] and [R] are true and [R] is the correct reason for [A]. (B) Both [A] and [R] are true but [R] is not the correct reason for [A]. (C) Both [A] and [R] are false.

27 7 (D)[A] is false and [R] is true. (D) Sol. The approximate load flow solution emerges from fast decoupled load flow method not only due to several assumptions but also due to power mismatch vector tolerance. In fact the power mismatch vector tolerance decides the accuracy of result of load flow solution. Q.4 The op-amp shown in the figure has a finite gain A = and an infinite input resistance. A stepvoltage V mv is applied at the input at time t = as shown. Assuming that the operational amplifier i is not saturated, the time constant (in millisecond) of the output voltage V is (A) (B) (C) (D) (A) Sol. Given : A V V V NI Also VNI I V V A Applying KCL at inverting terminal VI Vi VI V R Cs I (i) Vi VI Cs VCs R R Vi VRCs VI RCs RCs From equation (i)

28 8 V Vi VRCs A RCs RCs RCs V V i RCs A RCs ARCs RCs V V i ( RCs) A RCs V A V ( A) RCs i Which is comparable to V V RCs i Which is complete to V A V s i Time constant RC msec 3 6 Hence, the correct option is (A) Q.4 Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below: C ( P).P 3 P ; MW P 5 MW C ( P ).5P P ; MW P 8 MW The incremental cost (in Rs/MWh) of the power plant when it supplies MW is. () Sol. Given : C ( P).P 3P, MW P 5 MW C( P ).5P P, MW P 8 MW Incremental cost of generators must be equal for economic dispatch of power I I C C d dp d dp C C. P 3. P. P. P (i) Also, PP Solving equation (i) and (ii) we get P, P But MW P 5 MW and MW P 8 MW P and P is not possible (ii)

29 9 Thus, P MW and P MW Now, I I I. C C C Rs/MWh Q.4 A solution of the ordinary differential equation d y dy 5 6y is such that y() and dt dt 3e dy y(). The value of () is. 3 e dt ( 3) Q.43 An 8-bit, unipolar Successive Approximation Register type ADC is used to convert 3.5 V to digital equivalent output. The reference voltage is +5 V. The output of the ADC at the end of 3 rd clock pulse after the start of conversion, is (A) (B) (C) (D) Sol. (A) Given 8-bit unipolar successive Approximation Register Reference voltage Vref 5V At start output is MSB is set to DAC output 5.5 Analog input Va V, MSB retained At end of clock output Next MSB is set to DAC output V 4 V 3.5 V 3.75 V, the second MSB is reset to a At end of clock 3 output Continues till 8 th clock end. Hence, the correct option is (A) Q.44 The figure shows a digital circuit constructed using negative edge triggered J-K flip flops. Assume a starting state of QQ Q will repeat after number of cycles of the clock CLK. (6) Sol. State table Present State Flip Flop input Next state Q Q J Q K J Q K Q Q

30 3 State diagram Since, J K, flip flop Q goes on toggling on each clock starting from. The output Q is used as clock for the synchronous counter consisting of two flip flops with output Q and Q. The frequency of output Q is half that of frequency of clock signal (initial clock) The starting of QQ Q will repeat after 6 cycle of clock (CLK) Q.45 A sustained three-phase fault occurs in the power system shown in the figure. The current and voltage phasors during the fault (on a common reference), after the natural transients have died down, are also shown. Where is the fault located? (A) Location P (B) Location Q (C) Location R (D) Location S

31 3 Sol. (B) A the fault location, the magnitude of current shoots and voltage dips. From the voltage and current phasor, I is maximum and V is minimum. Therefore the fault is located at Q. Hence, the correct option is (B) Q.46 The maximum value of ' a ' such that the matrix eigenvectors is 3 a has three linearly independent real (A) (B) (C) 3 (D) (B) Q.47 A self-commutating switch SW, operated at duty cycle δ is used to control the load voltage as shown in the figure. Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are (A) VL and (C) VL and (A) Sol. Given : V V C C Vdc (B) V Vdc (D) V L L Vdc and V Vdc and V C C V V dc dc Case I : When SW is ON for period T ON V L di L dt i.e. VL Positive

32 3 di I V L L dt T dc dc ON I ON V T (i) L Case II : When SW is OFF for period T OFF V L di L dt i.e. VL negative VC VL V dc ; V VC di L V dt From equation (i) V dc T dc ON VC L Vdc LTOFF T VC V V ON dc TOFF T T ON OFF V Vdc T TON V dc V T V T TON V dc where T ON T Duty ratio The voltage across inductor is positive during TON and negative during T OFF. The average value across inductor, VL Hence, the correct option is (A) Q.48 A separately excited DC motor runs at rpm on no load when its armature terminals are connected at a V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is Ω. The no-load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 5 rpm. When the load torque is reduced to 5% of the full load value under the same armature voltage conditions, the speed rises to 5 rpm. Neglecting the rotational losses, the full load armature current (in Ampere) is. (8) Sol. Given : A separately excited DC motor N No load speed rmp

33 33 Ra Armature resistance Ia Full load armature current Case I : At No load We know back emf Eb kn Where, k constant, Flux /pole, N Speed I No load armature current A E kn V I R b k k. Case II : At full load torque N 5rpm a (i) Eb kn V IaRa.5 V I a V I a Case III : At 5% of full load torque Armature current.5 I a Speed N 5rpm E kn V.5 I R b a a V.5 I a. 5 (ii) V.5I a 4 (iii) Solving equation (ii) and (iii) we get full load armature current I 8A a

34 34 Q.49 In the given circuit, the parameter k is positive, and the power dissipated in the Ω resistor is.5 W. The value of k is. (.5) Sol. Given : Power dissipated in the resistor.5 W Applying KCL at node A V kv 5 k V 5 5 k V (i) Power dissipated in resistor.5 V = Current through Voltage across I V V.5 V 5V Putting V 5 in equation (i), we get V k.5 Q.5 The circuit shown is meant to supply a resistive load RL from two separate DC voltage sources. The switches S and S are controlled so that only one of them is ON at any instant. S is turned on for. ms and S is turned on for.3 ms in a.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the output voltage V (in Volt) across R L is.

35 35 (7) Sol. Given : Case I : When S is ON for. m sec V V for. m sec Case II : When S is ON for.3 m sec V 5Vfor.3 m sec Output voltage waveform

36 36 T Average output voltage V( avg ) Vdt T (area under curve) T V =7 V ( avg ) 3 3 (..3 5) 3 Q.5 In the signal flow diagram given in the figure, u and u are possible inputs whereas y and y are possible outputs. When would the SISO system derived from this diagram be controllable and observable? (A) When u is the only input and y is the only output (B) When u is the only input and y is the only output (C) When u is the only input and y is the only output (D) When u is the only input and y is the only output (B)

37 37 Q.5 The transfer function of a second order real system with a perfectly flat magnitude response of unity has a pole at ( j3). List all the poles and zeros. (A) Poles at ( ± j3), no zeros. (B) Poles at (± j3), one zero at origin. (C) Poles at ( j3), ( + j3), zeroes at ( j3), ( + j3). (D) Poles at ( ± j3), zeros at ( ± j3). (D) Q.53 In a linear two-port network, when V is applied to Port, a current of 4 A flows through Port when it is short-circuited. When 5 V is applied to Port, a current of.5 A flows through a Ω resistance connected across Port. When 3 V is applied to Port, the current (in Ampere) through a Ω resistance connected across Port is. (.54) Sol. Given : The short circuit admittance parameter model of a two port network is I Y V Y V I YV YV Case I : V V, I 4 A, V (short circuit) From equation (ii) 4 Y.4 S Case II : V 5 V, I.5 A V.5 V From equation (ii) Y Y.6 S Case III : V 3 A, V I From equation (i) I YV YV I.43 (.6) I (i) (ii) I.545 A Q.54 The open loop poles of a third order unity feedback system are at,,. Let the frequency corresponding to the point where the root locus of the system transits to unstable region be K. Now suppose we introduce a zero in the open loop transfer function at 3, while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits to unstable region? (A) It corresponds to a frequency greater than Ks

38 38 (B) It corresponds to a frequency less than K (C) It corresponds to a frequency K (D) Root locus of modified system never transits to unstable region (D) Q.55 The circuit shown in the figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by v( t) sin t. If the circuit is in steady state. Then the RMS value of the current (in Ampere) flowing in the circuit is. ()

39 Q. A circular turn of radius m revolves at 6 rpm about its diameter aligned with the x-axis as shown in the figure. The value of is 4 7 in SI unit. If a uniform magnetic field intensity applied, then the peak value of the induced voltage, V turn (in Volts), is. H 7 zˆ A/m is (39.47) Q. The synchronous generator shown in the figure is supplying active power to an infinite bus via two short, lossless transmission lines, and is initially in steady state. The mechanical power input to the generator and the voltage magnitude E are constant. If one line is tripped at time t by opening the circuit breakers at the two ends (although there is no fault), then it is seen that the generator undergoes a stable transient. Which one of the following waveforms of the rotor angle δ shows the transient correctly? (A) (B) (C) (D) (A)

40 Sol. If one line is tripped at time t by opening the circuit breaker at the two ends, then it is seen that the generator undergoes a stable transient. Generator settles to a new final steady state speed with new rotor angle slightly larger than the previous while undergoing stable transient. Therefore, rotor angle swings a round its equilibrium position to finally settle at stable condition. Hence the correct option is (A). Q.3 The figure shows the per-phase equivalent circuit of a two-pole three-phase induction motor operating at 5 Hz. The air-gap voltage, V g across the magnetizing inductance, is V RMS and the slip S is.5. The torque (in Nm) produced by the motor is. (4.43) Sol. Given : The per phase equivalent circuit of a two-pole three-phase induction motor s.5, f 5 Hz N Rotor current, I ' I s f 5 3 P ' V g r ( x ') s.5 (.).5 I ' 5.9 A Torque produced by motor r ' 3 T ( I ') Nm s where, N s 6 s.5 3 T (5.9) s

41 3 T 4.66 Nm Q.4 A 4-pole, separately excited, wave wound DC machine with negligible armature resistance is rated for 3 V and 5 kw at a speed of rpm. If the same armature coils are reconnected to form a lap winding, what is the rated voltage (in volts) and power (in kw) respectively at rpm of the reconnected machine if the field circuit is left unchanged? (A) 3 and 5 (B) 5 and 5 (C) 5 and.5 (D) 3 and.5 (B) Sol. Given : P 4, V 3 V, P 5 kw, N rpm Separately excited wave wound DC machine Neglecting Case I : Wave wound R R a R, number of parallel path, A a Case II : Lap wound a PNZ E E 6A A Number of parallel path, A number of poles A P 4 E A Now E A E E 4 E E Also since Ra E V Now power V V.(i) P I a VI a P V I a V

42 4 V I V a a Ia I a I.(ii) From equation (i) and (ii), we have in lap winding number of parallel paths are doubled, as compared to wave winding and therefore total armature current becomes double. The power of the machine remains same. The machine with lap wound armature coils. Will have rated voltage And power V 3 5 V V P 5 kw Hence the correct option is (B). Q.5 Find the transformer ratios a and b such that the impedance ( Z in) is resistive and equals.5 Ω when the network is excited with a sine wave voltage of angular frequency of 5 rad/s. (A) a.5, b. (B) a., b.5 (C) a., b. (D) a4., b.5 (B) Sol. Given : 5 rad/sec, Z.5 C F L mh R.5 in Now X j L j j XC 3 L 5 5 j 6 jc j5

43 5 in Z.5 j5 j a b b.(i) Also Zin.5.(ii) Equating imaginary and real parts from equation (i) and (ii), we have j5 j b b 4 b.5.5 and a a a Hence the correct option is (B). 4.5 Q.6 In the following circuit, the input voltage V in is sin( t). For RC 5, the average voltage across R (in Volts) under steady-state is nearest to

44 6 (A) (B) 3.8 (C) (D) 63.6 (C) Sol. Given : V V sin wt sin ( t) in m Case I : For position half cycle of supply, end a is positive wrt end b and capacitor C charger to V m almost instantly with zero time constant through diode D diode D is forward biased. Case II : For negative half cycle of supply end a is negative wrt end b and capacitor C charges to V m almost instantly with zero time constant through diode D diode D is forward biased. Thus, the average voltage across R under steady state is almost V m i.e. V Since RC time constant Which is much larger than time period of very little when corresponding diode is off. Hence the correct option is (C). 5.6 sec Vin i.e. T.sec, f 5 the capacitor discharges Q.7 Match the following Instrument Types Used for P. Permanent magnet moving coil. DC only Q. Moving iron connected through. AC only Current transformer

45 7 (C) R. Rectifier 3. AC and DC S. Electrodynamometer P Q R S (A) 3 (B) 3 (C) 3 3 (D) 3 Sol.. Permanent magnet moving coil instrument : It is used for DC only.. Moving iron : It can based for both AC and DC but when connected through Current Transformer (CT) it is used for AC only. 3. The rectifier and electrodynamometer can be used for AC and DC both. P-, Q-, R-3, S-3 Q.8 Given f ( z) g( z) h( z), where f, g, h are complex valued functions of a complex variable z. Which (B) one of the following statements is TRUE? (A) If f( z) is differentiable at z, then gz ( ) and hz ( ) are also differentiable at z. (B) If gz ( ) and hzare ( ) differentiable at z, then f( z) is also differentiable at z. (C) If f( z) is continuous at z, then it is differentiable at z. (D) If f( z ) is differentiable at z, then so are its real and imaginary parts. Q.9 The current i (in Ampere) in the Ω resistor of the given network is. () Sol. Given :

46 8 The given figure can be redrawn as Since the bride is in balanced condition, the current i in the resistor is zero i A Q. In the following circuit, the transistor is in active mode and VC V. To get VC 4V, we replace RC with R '. C Then the ratio R R ' C C is. Sol. Given :.74 to.76 Applying KVL in input loop, we get I B VBE R B

47 9 And also IC I B Change in RC does not affect the base current I B and so collector current I C Thus collector current I C constant Now, applying KVL in output loop, we get Since for R, V V C C I C V V R C To get VC 4 V, R C is replaced by Now I C constant C 4 R ' R C RC ' 6 R 8 C C R C ' RC '.75 R C Q. The operational amplifier shown in the figure is ideal. The input voltage (in Volt) is Vi sin( t).the amplitude of the output voltage V (in Volt) is. (. to.4) Sol. Given : Z k 3 R

48 Z R XC Z R R 3 6 jc j jc j 4. Z j487.3 Z Since the given figure is in inverting amplifier mode V V i Z Z V V 3 i V sin ( t ) 3 V.45sin( t 5.48 ) V Amplitude of output voltage (in volts) is.45 Q. We have a set of 3 linear equations in 3 unknowns. X Y means X and Y are equivalent statements and X Y means X and Y are not equivalent statements. P : There is a unique solution. Q: The equations are linearly independent. R : All Eigen-values of the coefficient matrix are nonzero. S : The determinant of the coefficient matrix is nonzero. Which one of the following is TRUE? (A) P Q R S (B) P R Q S (C) P Q R S (D) P Q R S (A) Q.3 A series RL circuit is excited at t = by closing a switch as shown in the figure. Assuming zero initial conditions, the value of di dt at t is (A) V L (B) V R RV (C) (D) L

49 Sol. (D) Given : Taking Laplace of the given circuit V IS ( ) s R Ls V I( S) R Lss L V A B IS ( ) R s R Lss s L L Using partial fraction A si() s s V A R Ls L V A R s R And B s I() s L V B Ls R s L R s L V B R V V IS ( ) Rs R Rs L V IS ( ) R s R s L

50 Taking inverse Laplace, we get Differentiating equation (i) wrt t V i( t) e R di() t V e dt L Again differentiating above wrt t Now d i t Rt L Rt L () VR e dt L d i t () dt Hence the correct option is (D). t VR L Rt L Q.4 A 3-bus power system network consists of 3 transmission lines. The bus admittance matrix of the uncompensated system is j6 j3 j4 j3 j7 j5 pu. j4 j5 j8.(i) If the shunt capacitance of all transmission lines is 5% compensated, the imaginary part of the 3 rd 3 rd column element (in pu) of the bus admittance matrix after compensation is (A) j7. (B) j8.5 (C) j7.5 (D) j9. (B) Sol. Given : j6 j3 j4 Y j3 j7 j5 pu j4 j5 j8 row Also we know Y Y Y3 Y Y3 Ybus Y Y Y Y3 Y3 Y Y Y Y Y Comparing given bus admittance matrix with its general form

51 3 Y3 Y3 j 4 Y3 Y3 j 5 Y3 j Y3 Y3 Y3 j8 Y3 ( j 4) ( j 5) j8 After 5% compensation in shunt capacitance Y 3 changes to Y '.5 Y 3 3 Y '.5 j 3 Y '.5 j 3 The 3 rd row and 3 rd column element of matrix Y after compensation will be Hence the correct option is (B). Y ' Y Y j.5 ( j 4) ( j 5) Y ' Y Y j Q.5 A shunt-connected DC motor operates at its rated terminal voltage. Its no-load speed is radian/second. At its rated torque of 5 Nm, its speed is 8 radian/second. The motor is used to directly drive a load whose load torque T L depends on its rotational speed r (in radian/second), such that T.78. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, L when it drives this load, is. (77 to 83) Sol. Given : No load speed, rad/sec At rated torque, r T 5 Nm rm 8 radian/sec Speed torque graph of the motor is shown as rm r (speed (rad/sec)) Using equation of straight line, Torque at any speed, rad/sec can be written as T m Also load torque T.78 L r 5 Nm r

52 4 The speed of the motor becomes constant when motor torque is equal to load torque i.e. r 5.78 r 8 rad/sec r Q.6 Match the following A dv A ds P. Stoke s Theorem. Dds Q Q. Gauss s Theorem. f ( z) dz R. Divergence Theorem 3. ( ) Ads S. Cauchy s Integral Theorem 4. ( ) Adl P Q R S (A) 4 3 (B) 4 3 (C) 4 3 (D) 3 4 (B) Q.7 The Laplace transform of f( t) t 3 is s (A) (B) 5 s 3 t Sol. Given : f( t), Now F() s (B) 3 s gt () s f( t) t t f() t L[ g( t)] G( s) L t F() s ds S 3 s ds S 3 s 3 S. The Laplace transform of (C) s gt () is t (D) 3 s

53 5 G() s Hence the correct option is (B). s Q.8 Two semi-infinite dielectric regions are separated by a plane boundary at y =. The dielectric constants of region (y < ) and region (y > ) are and 5, respectively. Region has uniform electric field E 3aˆ 4 ˆ ˆ x ay az, where aˆ, aˆ and a ˆz are unit vectors along the x, y and z axes, respectively. The electric field in region is x y (A) 3aˆ.6aˆ aˆ (B).aˆ 4aˆ aˆ x y z x y z (C).aˆ 4aˆ.8aˆ (D) 3aˆ aˆ.8aˆ (A) x y z x y z Q.9 When a bipolar junction transistor is operating in the saturation mode, which one of the following statements is TRUE about the state of its collector-base (CB) and the base-emitter (BE) junctions? (C) Sol. (A) The CB junctions is forward biased and the BE junction is reverse biased. (B) The CB junctions is reverse biased and the BE junction is forward biased. (C) Both the CB and BE junctions are forward biased (D) Both the CB and BE junctions are reverse biased FB forward biased, RB reversed biased (BE) Junction (CB) Junction Operating mode (i) FB RB Active mode (ii) RB RB Cut off mode (iii) FB FB Saturation mode (iv) RB RB Inverse active Hence the correct option is (C). Q. Consider the following Sum of Products expression, F. F ABC ABC ABC ABC ABC (A) The equivalent Product of Sums expression is (A) F ( A B C)( A B C)( A B C) (B) F ( A B C)( A B C)( A B C) (C) F ( A B C)( A B C)( A B C) (D) F ( A B C)( A B C)( A B C) Sol. Given : F ABC ABC ABC ABC ABC

54 6 The POS expression is F ( A B C)( A B C)( A B C) Hence the correct option is (A). Q. The filters F and F having characteristics as shown in figure (a) and (b) are connected as shown in figure (c). The cut-off frequencies of F and F are f and f respectively. If f f the characteristic of a, the resultant circuit exhibits (A) Band-pass filter (B) Band-stop filter (C) All pass filter (D) High-Q filter (A) Sol. Given : The filter F and f are low pass filter and high pass filter respectively according to their characteristics. When F and F are connected as shown below, will together realize a band stop filter provided the cut off frequencies of F and F respectively are f and f and f f

55 7 The realization of band stop filter is demonstrated below Hence the correct option is (B). Q. A 3-phase balanced load which has a power factor of.77 is connected to a balanced supply. The power consumed by the load is 5 kw. The power is measured by the two-wattmeter method. The readings of the two wattmeters are (A) Sol. (A) 3.94 kw and.6 kw (C) 5. kw and. kw (A) Given : In two wattmeter method Power factor, cos.77 Power factor angle, Power consumed by load, We know, power factor angle cos (.77) 45 (B).5 kw and.5 kw (D).96 kw and.4 kw ww 5kW.(i) V3 ( w w ) tan w w

56 8 tan 45 Solving equation (i) and (ii) we get Hence the correct option is (A). V ( w w ) 5 3 3( w w ) 5.(ii) w 3.94 kw w.6 kw Q.3 Nyquist plots of two functions G () s and G () s are shown in figure Nyquist plot of the product of G () s and G () s is (A) (B) (C) (D)

57 9 (B) Q.4 An open loop control system results in a response of (.4) DC gain of the control system is. t e (sin5t cos5 t) for a unit impulse input. The Q.5 A capacitive voltage divider is used to measure the bus voltage V bus in a high-voltage 5 Hz AC system as shown in the figure. The measurement capacitors C and C have tolerances of ±% on their nominal capacitance values. If the bus voltage V bus is kv RMS, the maximum RMS output voltage V out (in kv), considering the capacitor tolerance, is. (.75 to.5) Sol. Given : V kv Bus C F % Cmax. F and Cmin.9 F and C 9F % Cmax 9.9 F and Cmin 8. F Now from figure V out V bus C C C For Vout to be maximum, C must be maximum and ( C C) must be minimum

58 V out max out max Cmax. V bus C C.9 8. min min V. kv Q.6 Two semi-infinite conducting sheets are placed at right angles to each other as shown in the figure. A point charge of + Q is placed at a distance of d from both sheets. The net force on the charge is Q 4 K, where K is given by d (A) (B) iˆ ˆj 4 4 (C) iˆ ˆj (D) ˆ i ˆj (D) Q.7 In the given network Ampere) is V V, V V, V V. The phasor current i (in 3 (A) (B) (A) Sol. Given : V, V, The given network can be redrawn as 73. (C) V3. 6 (D).

59 Applying KCL at node a, we get V V V V j j 3 3 i V V V V i j j 3 3 Hence the correct option is (A). Q.8 Consider a signal defined by (A) Its Fourier Transform is sin( ) (A) (C) sin Sol. Given : ( ) ( ) ( ) ( ) i j j i jt e for t xt () for t Fourier transform of xt () is jt e for t xt () for t jt X ( ) x( t) e dt jt jt X ( ) e e dt j sin( ) (B) e j sin (D) e X ( ) j( ) t e j( )

60 e X ( ) X ( ) e j( ) j( ) j( ) j( ) j( ) ( e e ) j( ) sin ( ) X ( ) Hence the correct option is (A). Q.9 Two three-phase transformers are realized using single-phase transformers as shown in the figure. The phase difference (in degree) between voltage V and V is. ( 3 leading) Sol. Phase sequence assumed to be ABC Phasor diagram for delta/delta connected transformer

61 3 Phasor diagram for delta/delta connected transformer It can be seen that phase difference between V and V is 3. V is infect leading V by Q.3 The following discrete-time equations result from the numerical integration of the differential equations of an un-damped single harmonic oscillator with state variables x and y. The integration time step is h. x x k k y h y y k k xk h k For this discrete-time system, which one of the following statements is TRUE? (A) The system is not stable for h (B) The system is stable for (C) The system is stable for (D) The system is stable for (A) Sol. Given : and x h h h x y k k k or k k k y h y x x x hy k k k or k k k The z-transform of above discrete time equations gives h y Y hx 3. zx ( z) X ( z) hy ( z).(i) and zy( z) Y( z) hx ( z).(ii) from equation (i) we have Putting it in equation (ii) we get ( z ) Y ( z) X ( z) h ( z)( z) X ( z) hx ( z) h

62 4 ( z) h The oscillator in this question has the characteristic equation ( z ) h or z z h and characteristic roots are z: z jh. Since for all h, the roots will be outside unit circle on z- plane, the system is not stable for h Hence the correct option is (A). Q.3 For the switching converter shown in the following figure, assume steady-state operation. Also assume that the components are ideal, the inductor current is always positive and continuous and switching period is T s. If the voltage V L is as shown, the duty cycle of the switch S is. (.75) Sol. Given : Step up converter Case I : When switch S is ON di VL L Vin 5V dt I L 5.(i) T ON Case II : When switch S is OFF

63 5 di I VL L L 45 dt T OFF I L 45.(ii) T OFF Dividing equation (i) by (ii) Adding both sides duty ratio T T T T T OFF ON OFF ON OFF I 45 T T ON ON 4 3 TON 3.75 T T 4 OFF ON Q.3 With an armature voltage of V and rated field winding voltage, the speed of a separately excited DC motor driving a fan is rpm, and its armature current is A. The armature resistance is Ω. The load torque of the fan load is proportional to the square of the rotor speed. Neglecting rotational losses, the value of the armature voltage (in Volt) which will reduce the rotor speed to 5 rpm is. (47.5) Sol. Given : Separately exited DC motor At N rpm, I A, R Also load torque TL a N a Let the torque at N rpm be T and N 5 rpm be T T T N N 5 T T 4 (i) Also load torque is proportional to product of flux and armature current. Flux being constant

64 6 From equation (i) T T I I I I a a a a 4 Ia.5 A 4 Now, Eb kwm V IaRa V IaRa Speed k At N rpm, At N 5 rpm, k.5 5 V k Dividing equation (ii) by (iii) we get 5 V.5 V 47.5 V (ii) (iii) Q.33 Two coins R and S are tossed. The 4 joint events HRHS, TRTS, H RTS, TRH S have probabilities.8,.8,.3,.4 respectively where H represents head and T represents tail. Which one of the following is TRUE? (A) The coin tosses are independent (B) R is fair, S is not (C) S is fair, R is not (D) The coin tosses are dependent (D) Q.34 A composite conductor consists of three conductors of radius R each. The conductors are arranged as shown below. The geometric mean radius (GMR) (in cm) of the composite conductor is kr. The value of k is. (.94) Sol. Given :

65 7 The geometric mean radius (GMR) of composite conductor 3 9 KR [ AB. AC. BA. BC. CACB..(.7788 R) ] [(3 R).7788 R) ] K.93 Q.35 An open loop transfer function G(s) of a system is Sol. K Gs () s( s )( s ) For a unity feedback system, the breakaway point of the root loci on the real axis occurs at, (A).4 (B).58 (C).4 and.58 (A) (D) None of the above Q.36 Two identical coils each having inductance L are placed together on the same core. If an overall inductance of L is obtained by interconnecting these two coils, the minimum value of is. () Sol. The connection of two identical coils, each with inductor L, to result in minimum value of overall inductance L is Where M mutual inductance For L L L L M ( L M) L to be minimum, M must be maximum Now M k L L eq M kl For M to be maximum coupling coefficient k should be i.e. k M L Now L ( L M )

66 8 ( L L) L min Q.37 A three-winding transformer is connected to an AC voltage source as shown in the figure. The number of turns are as follows : N, N 5, N3 5. If the magnetizing current is neglected and the currents in two windings are in Ampere? I 3 A and I3 5 A, then what is the value of the current I (A) (A) Sol. Given : 9 (B) 7 (C) 4 9 (D) 4 7 N, N 5 N, 3 5 I 3 A, Primary mmf = secondary mmf + tertiary mmf Hence the correct option is (A). IN IN I3N3 N I I I 3 3 N N I3 5 A N I 9 A

67 9 Q.38 The saturation voltage of the ideal op-amp shown below is ± V. The output voltage v of the following circuit in the steady-state is (A) Square wave of period.55 ms (C) Square wave of period.5 ms (A) Sol. Given : (B) Triangular wave of period.55 ms (D) Triangular wave of period.5 ms The given circuit is astable multivibrator whose output is a square wave of time period ( T ). where, ( ) T RC ln ( ) feedback factor T.5 ln.5 T.55 m sec Q.39 The coils of a wattmeter have resistance. Ω and Ω; their inductances may be neglected. The wattmeter is connected as shown in the figure, to measure the power consumed by a load, which draws

68 3 5 A at power factor.8. The voltage across the load terminals is 3 V. The percentage error on the wattmeter reading is. (.5) Sol. Given : R, R. P C The pressure coil is connected at the load end The power measured by wattmeter is the sum of power consumed by load and power loss in pressure coil. Measured power Pw PPC V VI cos R P True power % error 6.9 W VI cos W Measured power True power True power %

69 3 4 3 Q.4 The z-transform of a sequence xn [ ] is given as X ( z) z 4 z z. If yn [ ] is the first difference of xn [ ], then Yx [ ] is given by (A) z (B) z 3 3 z z z z z z (C) z (D) 4z 3 3 z z z z z z (A) Sol. The z-transform of a sequence Xn [ ] is 4 3 Given : X ( z) z 4 z z The first difference of xn [ ] is Yn [ ] Y[ n] x[ n] x[ n ] Taking z-transform both side, we have Hence the correct option is (A). Y z x Z Z X z ( ) ( ) ( ) z 4 z 4 z Z Z z z z z Z Z 3 Q.4 A balanced (positive sequence) three-phase AC voltage source is connected to a balanced, star connected load through a star-delta transformer as shown in figure. The line-to-line voltage rating is 3 V on the star side, and 5 V on the delta side. If the magnetizing current is neglected and I s A, then what is the value on p I in Ampere?

70 3 Sol. (A) (A) 5 3 (B) 5 3 (C) (D) 3 Assuming phase sequence ABC, the phasor diagram on primary and secondary side are drawn Is A (given) (line current) Is (phase) 3 A Primary side current I p is in phase with AA voltage which is the phase voltage. The line voltage V L is CA The I p will be leading its line voltage V L by Now primary phase voltage V p, phase 3 V 3 Secondary phase voltage V s, phase 5V. The currents are inversely proportional to voltage Hence the correct option is (A). Ip(phase) V I (phase) V s 5 I p s,phase p,phase I p A leading the line voltage by 3 I p 5 3 Q.4 A buck converter feeding a variable resistive load is shown in the figure. The switching frequency of the switch S is khz and the duty ratio is.6. The output voltage V is 36 V. Assume that all the components are ideal, and that the output voltage is ripple-free. The value of R (in Ohm) that will make the inductor current ( i L ) just continuous is. 3

71 33 (48 to 5) Sol. Given ; A buck converter Current waveform TON f khz, duty ratio.6 T V 36 V L 5 mh I I I max I max I I max During T OFF period when switch is OFF V di L dt min 3 max min Imin TOFF 36 5 I I 36T OFF max ( ) 3 5 f 36 (.6) 5 max I 3 3 I.88 A. For continuous conduction of inductor current